How do you solve $x - \frac{4}{x} - 1 = \frac{12}{3} - x + 1$ $x-4/x-1 = 12/3-x +1$ ?

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solving rational equations

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Answer 1 · 2018-07-20T04:29:28

$x = \frac{3 + \sqrt{17}}{2}$ $x=(3+sqrt17)/2$ or $x = \frac{3 - \sqrt{17}}{2}$ $x=(3-sqrt17)/2$

$x - \frac{4}{x} - 1 = \frac{12}{3} - x + 1$ $x-4/x-1=12/3-x+1$ where $x \neq 0$ $x!=0$

$x - \frac{4}{x} - 1 = 4 - x + 1$ $x-4/x-1=4-x+1$

$2 x - 6 = \frac{4}{x}$ $2x-6=4/x$

$2 x^{2} - 6 x = 4$ $2x^2-6x=4$

$2 x^{2} - 6 x - 4 = 0$ $2x^2-6x-4=0$

$x^{2} - 3 x - 2 = 0$ $x^2-3x-2=0$

$x = \frac{3 \pm \sqrt{9 + 8}}{2}$ $x=(3+-sqrt(9+8))/2$

$x = \frac{3 \pm \sqrt{17}}{2}$ $x=(3+-sqrt17)/2$

$x = \frac{3 + \sqrt{17}}{2}$ $x=(3+sqrt17)/2$ or $x = \frac{3 - \sqrt{17}}{2}$ $x=(3-sqrt17)/2$